数据库:(gameDay)
id game_Date
1 1996-01-02自从这场比赛发生以来,我一直在努力归还这段时间。
我试图运行的sql查询是:
SELECT CURDATE() - gameDay.game_Date AS days
FROM gameDay
WHERE gameDay.id = 1这是给了我奇怪的数字,这是毫无意义的:
或
SELECT DATEDIFF(CURDATE(), gameDay.game_Date) AS days
FROM gameDay
WHERE gameDay.id = 1这个sql查询给了我一个错误
未定义的函数或表达式: CURDATE
或
选择DATEDIFF(DATEDIFF,CURDATE(),gameDay.game_Date)作为gameDay的天数,其中gameDay.id =1
这个查询给了我一个错误:
字段不是基于查询: DAY中的表的有效字段。
谢谢
发布于 2017-03-16 10:33:50
IF you use SQL Server database means try below query :
SELECT DATEDIFF(DAY,GETDATE(), gameDay.game_Date) AS days
FROM gameDay
WHERE gameDay.id = 1
IF you use MySQL database means try below query :
SELECT DATEDIFF(`game_Date`, NOW()) AS days
FROM `gameDay`
WHERE `id` = 1https://stackoverflow.com/questions/42831432
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